Problem: Baby Ehab was toying around with arrays. He has an array a of length n. He defines an array to be good if there's no way to partition it into 2 subsequences such that the sum of the elements in the first is equal to the sum of the elements in the second. Now he wants to remove the minimum number of elements in a so that it becomes a good array. Can you help him?

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly zero or all) elements. A partitioning of an array is a way to divide it into 2 subsequences such that every element belongs to exactly one subsequence, so you must use all the elements, and you can't share any elements.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 2000) — the elements of the array a.

Output

The first line should contain the minimum number of elements you need to remove.

The second line should contain the indices of the elements you're removing, separated by spaces.

We can show that an answer always exists. If there are multiple solutions, you can print any.

Examples

Input


4
6 3 9 12


Output


1
2

Input


2
1 2


Output


0

Note

In the first example, you can partition the array into [6,9] and [3,12], so you must remove at least 1 element. Removing 3 is sufficient.

In the second example, the array is already good, so you don't need to remove any elements.

Solution (in PYTHON3):

Solution:
import sys,heapq,math
from collections import defaultdict,deque
input=sys.stdin.readline

def check(arr,su):
    su=su//2
    dp=[[False for _ in range(su+1)] for _ in range(n+1)]
    for i in range(n+1):   #col->sum,row->index
        dp[i][0]=True
    for i in range(1,n+1):
        for j in range(1,su+1):
            dp[i][j]=dp[i-1][j]
            if(j>=arr[i-1]):
                dp[i][j]=dp[i][j] or dp[i-1][j-arr[i-1]]
    return dp[-1][-1]
    

n=int(input())
l=list(map(int,input().split()))
su=sum(l)
if(su%2==1 or not check(l,su)):
    print(0)
else:
    #can be partitioned
    #sum is even
    f=0
    for i in range(n):
        if(l[i]%2==1):  #found odd element,remove it
            print(1) #even-odd=odd,it cant be partitioned now
            print(i+1)
            f=1
            break
    if(f==0): #means all even elements
        '''
        s={a,b,c,d,e}
        s1={a,b,c} s2={d,e}   a+b+c=d+e
        if set s can be partitioned into 2 sets s1,s2,
        if we multiply all the elements by 2,
        s1={2a,2b,2c},s2={2d,2e}
        sum(s1)=2(a+b+c)
        sum(s2)=2(d+e)
        since a+b+c==d+e, ==> s1=s2

    32 24 56 (original array)
    16 12 28 
    8 6 14  
    4 3 7  (if this can be partitioned,all the above arrays(*2) can also be partitioned) (found odd element)


    suppose i remove an odd element(in last array) (now it cant be partitioned!!)

    4 7 (which cant be partitioned)
    even if i multiply the array by 2 any number of times,it cant be partitioned!!
    ex:
    4 7
    8  14
    16 28
    32 56 (we get original array with one element removed and it cant be partitioned now)

                                                  
        '''
        #keep on dividing the elements by 2,until you find an odd element!
        f=0
        while(1):
            for i in range(n):
                l[i]=l[i]//2
                if(l[i]%2!=0):
                    print(1)
                    print(i+1)
                    f=1
                    break
                
            if(f==1):
                break